What does permute actually produce?

Trying some calculations earlier, I have realised permute does not do what I thought it did. To clarify I did this

    % code
test = rand(4,4,4,4) + rand(4,4,4,4)*1i;
test2a = zeros(4,4,4,4); test2b = zeros(4,4,4,4); 
for a = 1:4
    for b = 1:4
        for c = 1:4
            for d = 1:4
test2a(a,b,c,d) = test(a,b,d,c);
test2b(a,b,c,d) = test(b,a,c,d); 
            end
        end
    end
end
test3a = permute(test,[1,2,4,3]);
test3b = permute(test,[2,1,3,4]);
m1=max(test2a(:)-test3a(:))
m2=max(test2b(:)-test3b(:))
test2a = zeros(4,4,4,4); test2b = zeros(4,4,4,4); 
for a = 1:4
    for b = 1:4
        for c = 1:4
            for d = 1:4
test2a(a,b,c,d) = test(d,b,a,c);
test2b(a,b,c,d) = test(c,a,b,d); 
            end
        end
    end
end
test3a = permute(test,[4,2,1,3]); 
test3b = permute(test,[3,1,2,4]);
m3=max(test2a(:)-test3a(:))
m4=max(test2b(:)-test3b(:))
m1 =

     0

m2 =

     0

m3 =

   0.9525 + 0.5183i

m4 =

   0.9293 - 0.7266i

clearly the first is the same but the latter is different. Is there some way to create the second variables (of form test2(a,b,c,d) == test(d,b,a,c) which involve two permutations ) just using permute on test?

[David H]

I realise reading this again I haven't really stated what I want to do very clearly.

If I want some variable satisfying var2(a,b,c,d) == var1(d,b,a,c) for any a b c d (up to the size of the variable) how do I produce this variable efficiently from var1?

Answer 1

[the cyclist]

Isn't it

var2 = permute(var1,[3 2 4 1]);

?

[David H]

It is indeed, for some reason I can't quite see why that is rather than var2 = permute(var1,[4,2,1,3]). Presumably due to the multiple transposes that have to occur. Is there an easy way to see this rule?

[the cyclist]

Here is the reasoning:

• The former dimension 3 is now dimension 1.
• The former dimension 2 is now dimension 2.
• The former dimension 4 is now dimension 3.
• The former dimension 1 is now dimension 4.

So, you see that new dimension [1 2 3 4] is old dimension [3 2 4 1].

That latter vector is the one you needed.

[David H]

Ah yes I was thinking on the lines of:

• The new dimension 1 is dimension 4,
• The new dimension 2 is dimension 2,
• new dimension 3 is dimension 1 etc

rather than this way round, which is of course not the values you call permute with!

Thanks for the clarification, I realise it was a fairly nieve mistake.